The sum of the roots of the equation `|sqrt3cos x-sinx|=2" in "[0, 4pi]` is `kpi`, then the value of 6k is

Answer

To solve the equation \( |\sqrt{3} \cos x - \sin x| = 2 \) in the interval \( [0, 4\pi] \) and find the value of \( 6k \) where the sum of the roots is \( k\pi \), we can follow these steps:

### Step 1: Set Up the Equation
We start with the equation:
\[ |\sqrt{3} \cos x - \sin x| = 2 \]
This absolute value equation can be split into two cases:
1. \( \sqrt{3} \cos x - \sin x = 2 \)
2. \( \sqrt{3} \cos x - \sin x = -2 \)

### Step 2: Solve the First Case
For the first case:
\[ \sqrt{3} \cos x - \sin x = 2 \]
Rearranging gives:
\[ \sqrt{3} \cos x = \sin x + 2 \]

### Step 3: Solve the Second Case
For the second case:
\[ \sqrt{3} \cos x - \sin x = -2 \]
Rearranging gives:
\[ \sqrt{3} \cos x = \sin x - 2 \]

### Step 4: Transform the Equations
We can rewrite the equations in terms of sine and cosine:
1. For the first case, divide by 2:
\[ \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x = 1 \]
Recognizing \( \frac{\sqrt{3}}{2} = \sin\left(\frac{\pi}{3}\right) \) and \( \frac{1}{2} = \cos\left(\frac{\pi}{3}\right) \), we can use the sine subtraction identity:
\[ \sin\left(\frac{\pi}{3} - x\right) = 1 \]

2. For the second case:
\[ \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x = -1 \]
Similarly, we have:
\[ \sin\left(\frac{\pi}{3} - x\right) = -1 \]

### Step 5: Find the Roots
From \( \sin\left(\frac{\pi}{3} - x\right) = 1 \):
\[ \frac{\pi}{3} - x = \frac{\pi}{2} + 2n\pi \quad \text{(for } n \in \mathbb{Z}\text{)} \]
This gives:
\[ x = \frac{\pi}{3} - \frac{\pi}{2} + 2n\pi = -\frac{\pi}{6} + 2n\pi \]
And:
\[ x = \frac{\pi}{3} - \frac{5\pi}{2} + 2n\pi = -\frac{13\pi}{6} + 2n\pi \]

From \( \sin\left(\frac{\pi}{3} - x\right) = -1 \):
\[ \frac{\pi}{3} - x = \frac{3\pi}{2} + 2n\pi \]
This gives:
\[ x = \frac{\pi}{3} - \frac{3\pi}{2} + 2n\pi = -\frac{7\pi}{6} + 2n\pi \]

### Step 6: Collect All Roots in the Interval
We need to find all roots in the interval \( [0, 4\pi] \):
1. For \( n = 0 \):
- \( x = -\frac{\pi}{6} \) (not in range)
- \( x = -\frac{13\pi}{6} \) (not in range)
- \( x = -\frac{7\pi}{6} \) (not in range)

2. For \( n = 1 \):
- \( x = \frac{11\pi}{6} \)
- \( x = \frac{5\pi}{6} \)
- \( x = \frac{23\pi}{6} \)
- \( x = \frac{17\pi}{6} \)

### Step 7: Calculate the Sum of the Roots
Now we sum the valid roots:
\[ \frac{5\pi}{6} + \frac{11\pi}{6} + \frac{17\pi}{6} + \frac{23\pi}{6} = \frac{5 + 11 + 17 + 23}{6}\pi = \frac{56}{6}\pi = \frac{28}{3}\pi \]
Thus, \( k = \frac{28}{3} \).

### Step 8: Find \( 6k \)
Finally, we calculate:
\[ 6k = 6 \times \frac{28}{3} = 56 \]

### Conclusion
The value of \( 6k \) is \( \boxed{56} \).